Q.71. 5.33 g of CrCl₃·6H₂O, which is a 1 : 3 electrolyte, is dissolved in water and is passed throug

Q: What is the correct answer for this exam Chemistry Q71?

The correct answer is 33. Sol. The complex CrCl₃·6H₂O is a 1:3 electrolyte. This means that when it dissolves, it produces one complex cation and three counter-anions. Since chloride ions are present both inside and outside the coordination sphere, and it's a 1:3 electrolyte, it must be [Cr(H₂O)₆]Cl₃. The three chloride ions are outside the coordination sphere and are ionizable. Reaction with AgNO₃: [Cr(H₂O)₆]Cl₃(aq) + 3A

Q: Q.71. 5.33 g of CrCl₃·6H₂O, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with AgNO₃, results in…

The correct answer is 33. Sol. The complex CrCl₃·6H₂O is a 1:3 electrolyte. This means that when it dissolves, it produces one complex cation and three counter-anions. Since chloride ions are present both inside and outside the coordination sphere, and it's a 1:3 electrolyte, it must be [Cr(H₂O)₆]Cl₃. The three chloride ions are outside the coordination sphere and are ionizable. Reaction with AgNO₃: [Cr(H₂O)₆]Cl₃(aq) + 3A

Question

Q.71. 5.33 g of CrCl₃·6H₂O, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with AgNO₃, results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is _________ . (Nearest integer)
[Molar mass in g mol⁻¹ Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

FirstInTest · Accepted Answer

Sol. The complex CrCl₃·6H₂O is a 1:3 electrolyte. This means that when it dissolves, it produces one complex cation and three counter-anions. Since chloride ions are present both inside and outside the coordination sphere, and it's a 1:3 electrolyte, it must be [Cr(H₂O)₆]Cl₃. The three chloride ions are outside the coordination sphere and are ionizable.

Reaction with AgNO₃:
[Cr(H₂O)₆]Cl₃(aq) + 3AgNO₃(aq) → [Cr(H₂O)₆](NO₃)₃(aq) + 3AgCl(s)
From the stoichiometry, 1 mole of the complex reacts to form 3 moles of AgCl.

Molar mass of AgCl = 108 (Ag) + 35.5 (Cl) = 143.5 g/mol
Moles of AgCl formed = $\frac{	ext{mass of AgCl}}{	ext{molar mass of AgCl}} = \frac{8.61 	ext{ g}}{143.5 	ext{ g/mol}} = 0.06 	ext{ mol}$

From the reaction stoichiometry, if 0.06 mol of AgCl is formed, then moles of complex reacted = $\frac{0.06 	ext{ mol}}{3} = 0.02 	ext{ mol}$

The question asks for the ratio of moles of complex reacted and moles of AgCl formed.
Ratio = $\frac{	ext{moles of complex reacted}}{	ext{moles of AgCl formed}} = \frac{0.02 	ext{ mol}}{0.06 	ext{ mol}} = \frac{1}{3}$

The answer needs to be in the format $	imes 10^{-2}$.
$\frac{1}{3} = 0.3333...$
To express this as $	imes 10^{-2}$, we multiply by 100:
$0.3333... 	imes 100 = 33.33...$

Rounding to the nearest integer, the value is 33.

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